# $\mathcal{PT}$ symmetry is a weaker condtion for a Hamiltonian than Hermicity?

In this paper by Bender (end of page 3), he says:

And, because PT-symmetry is a weaker condition than Hermiticity, there are infinitely many Hamiltonians that are PT-symmetric but non-Hermitian; we can now study new kinds of quantum theories that would have been rejected in the past as being unphysical.

Similar claims have been made in his former PRL papers (1998 and 2002), namely less restrictive, weaker condition and so on.

I am a bit confused. Since there are also many Hamiltonians that are Hermitian but have no PT symmetry, one should not say that PT-symmetry is a weaker condition. One example is a particle in an asymmetric real potential $$V(x)$$, which doesn't have PT-symmetry but is Hermitian.

• Maybe he meant that $\mathcal{P}\mathcal{T}$ is weaker than $\mathcal{P}\mathcal{T}$ + Hermiticity, that is, $\mathcal{P}\mathcal{T}$ is implicitly always considered ? I agree it seems rather strange. – Lucas Baldo May 7 at 4:44